dataframe - How to split columns in data frame by semicolon in R -

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my question feels me obvious, however, not find solution.

a have data frame this:

<ticker>;<per>;<date>;<time>;<open>;<high>;<low>;<close> usd index;d;20150801;000000;97.199;97.336;97.191;97.192 usd index;d;20150802;000000;97.226;97.294;97.207;97.257 usd index;d;20150803;000000;97.255;97.582;97.155;97.499

i need them split in different columns ; this:

<ticker>    <per>   <date>  <time>  <open>  <high>  <low>   <close> usd index   d   20150801    0   97.199  97.336  97.191  97.192 usd index   d   20150802    0   97.226  97.294  97.207  97.257 usd index   d   20150803    0   97.255  97.582  97.155  97.499

this basic question needs @ top of search results. thank in advance helping me resolve issue!

we can use read.table

setnames(read.table(text=dat[,1], sep=";", stringsasfactors=false),             scan(text=names(dat), sep=";", = "", quiet = true)) #    <ticker> <per>   <date> <time> <open> <high>  <low> <close> # 1 usd index     d 20150801      0 97.199 97.336 97.191  97.192 # 2 usd index     d 20150802      0 97.226 97.294 97.207  97.257 # 3 usd index     d 20150803      0 97.255 97.582 97.155  97.499

data

dat <- structure(list(`<ticker>;<per>;<date>;<time>;<open>;<high>;<low>;<close>` =  c("usd index;d;20150801;000000;97.199;97.336;97.191;97.192",  "usd index;d;20150802;000000;97.226;97.294;97.207;97.257",  "usd index;d;20150803;000000;97.255;97.582;97.155;97.499"  )), .names = "<ticker>;<per>;<date>;<time>;<open>;<high>;<low>;<close>",   class = "data.frame", row.names = c(na, -3l))

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